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babymother [125]
3 years ago
7

What is the answer x^2+28=0

Mathematics
2 answers:
marshall27 [118]3 years ago
5 0

Answer:

the answer is ( there is no real solutions) i hoped this help if you need an explanation why let me know :)

Blababa [14]3 years ago
3 0

Answer:

x = 2√7 or 5.2915

Step-by-step explanation:

x^2 + 28 = 0

first get the variable by itself by subtracting the 28 from both sides. 28-28 cancels and 0-28=-28.

now you have

x^2 = -28

take the square root of x^2 to cancel out the exponent, and then do the same to the other side,

x = √28

x = 2√7 or 5.2915

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What two numbers add up to -18 and multiply to -12?
gregori [183]

Answer:

-6

Step-by-step explanation:

6 0
3 years ago
Hi, I need help. The sheet is homework. If you help, thank you
Paul [167]

Answer:

1.  Solution

2.  Distributive Property

3.  Multiplying

4.  Area model

5.  Properties of equality

6.  Expanded form

Step-by-step explanation:

Hope this helps!

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7 0
3 years ago
Read 2 more answers
Solve the inequality for u.<br> 7u - 32&lt; - 3(6-3u)<br> Simplify your answer as much as possible.
Dmitry_Shevchenko [17]

Answer:

u > - 7

Step-by-step explanation:

7u - 32 < - 3(6-3u)

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18-32 < - 7u + 9u

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3 0
3 years ago
Read 2 more answers
(2a+b)^2 how do you do this
gulaghasi [49]

•  Expand (2a + b)²:

(2a + b)²

= (2a + b) · (2a + b)


Multiply out the brackets by applying the distributive property of multiplication:

= (2a + b) · 2a + (2a + b) · b

= 2a · 2a + b · 2a + 2a · b + b · b

= 2²a² + 2ab + 2ab + b²


Now, group like terms together, and you get

= 2²a² + 4ab + b²

= 4a² + 4ab + b²    <———    expanded form  (this is the answer).


I hope this helps. =)


Tags:  <em>special product square of a sum algebra</em>

3 0
3 years ago
How many numbers are there between 100,000 and 120,000 Where the digits are in order from least to greatest no digits can be rep
Fittoniya [83]

I don't really know what you mean by order from least to greatest but all I know is that there are 20,000 numbers in between 100,000 and 120,000.

7 0
4 years ago
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